Solution 9.21
(a)
oooo jF 86.343236.8758.48296.690304155 ∠=+=−−∠−∠=
(b)
ooo jG 49.62565.59358.4571.2504908 −∠=−=∠+−∠=
Solution 9.22
Let f(t) =
∫
−+
t
dttv
dt
dv
tv )(24)(10
Solution 9.23
(a) v = [110sin(20t+30˚) + 220cos(20t–90˚)] V leads to V = 110∠(30˚–90˚) + 220∠–90˚
Solution 9.24
(a)
1,010
j=ω°∠=
ω
+V
V
(b)
4),9010(20
j
4
5j =ω°−°∠=
ω
++ω V
VV
Solution 9.25
(a)
2,45432j =°∠=+
ωω
II
(b)
5,2256j
10 =ω°∠=+ω+
I
Solution 9.26
2,01
2j =ω°∠=
++ω I
II
Solution 9.27
377,10–110
j
10050j =ω°∠=
ω
++ω V
VV
Solution 9.28
Determine the current that flows through a 20-Ω resistor connected in parallel with a
Solution
Solution 9.29
Solution
5.0j–
)102)(10(j
1
Cj
1
6–6 =
×
=
ω
=Z
Solution 9.30
Since R and C are in parallel, they have the same voltage across them. For the resistor,
Solution 9.31
Solution
3
240 2 240 10 0.48L mH j L j x x j
ω
−
= → = =
Solution 9.32
Using Fig. 9.40, design a problem to help other students to better understand phasor
Problem
Two elements are connected in series as shown in Fig. 9.40.
If i = 12 cos (2t – 30°) A, find the element values.
Figure 9.40
Solution
V = 180∠10°, I = 12∠-30°, ω = 2
Solution 9.33
Solution
220∠θ = vR + jvL where 220 =
22
LR
vv +
Solution 9.34
Solution 9.35
Find current i in the circuit of Fig. 9.42, when vs(t) = 115cos(200t) V.
Figure 9.42
For Prob. 9.35.
Solution
3
11
5200 5 10
mF j
j C j xx
ω
−
→ = = −
Solution 9.36
Using Fig. 9.43, design a problem to help other students to better understand impedance.
Problem
In the circuit in Fig. 9.43, determine i. Let vs = 60 cos(200t – 10°) V.
Figure 9.43
Solution
Let Z be the input impedance at the source.
Solution 9.37
Determine the admittance Y for the circuit in Fig. 9.44.
Figure 9.44
For Prob. 9.37.
Solution
Let us start with Z = 1/Y = 10+j30 + (–j10)(10+j10)/(–j10+10+j10)
Solution 9.38
Using Fig. 9.45, design a problem to help other students to better understand admittance.
Problem
Find i(t) and v(t) in each of the circuits of Fig. 9.45.
Figure 9.45
Solution
12j)3)(4(jLjH3 ==ω→
Solution 9.39
For the circuit shown in Fig. 9.46, find Zeq and use that to find current I. Let ω=10 rad/s.
Figure 9.46
For Prob. 9.39.
Solution
Zeq = 4 + j20 + [16(–j14+j25)/(16–j14+j25)] = 4 + j20 + j176(16–j11)/(256+121)